the correct answer is option C) \(F\over\sqrt{mk}\)
Detailed solution::
Maximum speed is at mean position (equilibrium). F = kx
\(x={F\over k}\)
\(W_F + W_{sp} = \Delta KE
\)
\(F(x) – {1\over2} kx^2={1\over2}mv^2-0\)
\(F\left({F\over k}\right) – {1\over2} k\left({F\over k}\right)^2={1\over2}mv^2\)
\(\implies v_{max} = {F\over\sqrt{mk}}\)